Respuesta :
Answer:
H1: [tex] \sigma^2_1 \neq \sigma^2_2[/tex]
So then the correct option for this case would be:
H1:σ21≠σ22
Step-by-step explanation:
Data given and notation
[tex]n_1 = 10 [/tex] represent the sampe size for professor 1
[tex]n_2 =10[/tex] represent the sample size for professor 2
[tex]\bar X_1 =79.3[/tex] represent the sample mean for professor 1
[tex]\bar X_2 =82.1[/tex] represent the sample mean for professor 2
[tex]s_1 = 22.4[/tex] represent the sample deviation for professor 1
[tex]s^2_1 = 501.76[/tex] represent the sample variance for professor 1
[tex]s_2 = 12[/tex] represent the sample deviation for professor 2
[tex]s^2_2 = 144[/tex] represent the sample variance for professor 2
[tex]\alpha[/tex] represent the significance level provided
F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:
[tex]F=\frac{s^2_1}{s^2_2}[/tex]
Solution to the problem
System of hypothesis
We want to test if the variation between their grading procedure, so the system of hypothesis are:
H0: [tex] \sigma^2_1 = \sigma^2_2[/tex]
H1: [tex] \sigma^2_1 \neq \sigma^2_2[/tex]
So then the correct option for this case would be:
H1:σ21≠σ22
Calculate the statistic
Now we can calculate the statistic like this:
[tex]F=\frac{s^2_1}{s^2_2}=\frac{22.4^2}{12^2}=3.484[/tex]
Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have [tex]n_1 -1 =10-1=9[/tex] and for the denominator we have [tex]n_2 -1 =10-1=9[/tex] and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:
[tex]p_v =P(F_{9,9}>3.484)=0.0385[/tex]
And we can use the following excel code to find the p value:"=1-F.DIST(3.484,9,9;TRUE)"
Conclusion
Assuming [tex] \alpha=0.05[/tex] and since the [tex]p_v < \alpha[/tex] we have enough evidence to reject the null hypothesis.
